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The statement $(\mathrm{p} \wedge(\mathrm{p} \rightarrow \mathrm{q}) \wedge(\mathrm{q} \rightarrow \mathrm{r})) \rightarrow \mathrm{r}$ is :
a tautology
equivalent to $\mathrm{p} \rightarrow \sim \mathrm{r}$
a fallacy
equivalent to $\mathrm{q} \rightarrow \sim \mathrm{r}$
Solution
$(\mathrm{p} \wedge(\mathrm{p} \rightarrow \mathrm{q}) \wedge(\mathrm{q} \rightarrow \mathrm{r})) \rightarrow \mathrm{r}$
$\equiv(\mathrm{p} \wedge(\sim \mathrm{p} \vee \mathrm{q}) \vee(\sim \mathrm{q} \vee \mathrm{r})) \rightarrow \mathrm{r}$
$\equiv((\mathrm{p} \wedge \mathrm{q}) \wedge(\sim \mathrm{p} \vee \mathrm{r})) \rightarrow \mathrm{r}$
$\equiv(\mathrm{p} \wedge \mathrm{q} \wedge \mathrm{r}) \rightarrow \mathrm{r}$
$\equiv \sim(\mathrm{p} \wedge \mathrm{q} \wedge \mathrm{r}) \vee \mathrm{r}$
$\equiv(\sim \mathrm{p}) \vee(\sim \mathrm{q}) \vee(\sim \mathrm{r}) \vee \mathrm{r}$
$\Rightarrow \text { tautology }$
